Q.73. 20 g hemoglobin in a 1 L aqueous solution (A) at 300 K is separated from pure water by semi pe

Q: What is the correct answer for JEE Main Chemistry Q73?

The correct answer is 62. The osmotic pressure ($\pi$) can be calculated from the height difference ($h$) in the column: $\pi = \rho g h$ Given: $\rho$ (density of solution) = 1000 kg m$^{-3}$ $g$ = 10 m s$^{-2}$ $h$ = 80.0 mm = $80.0 \times 10^{-3}$ m $\pi = (1000 \text{ kg m}^{-3}) \times (10 \text{ m s}^{-2}) \times (80.0 \times 10^{-3} \text{ m})$ $\pi = 800 \text{ Pa}$ Now, use the van't Hoff equation for osmotic pr

Q: Q.73. 20 g hemoglobin in a 1 L aqueous solution (A) at 300 K is separated from pure water by semi permeable membrane. At equilibrium the height of solution in a tube dipped in a solution (A) is found…

The correct answer is 62. The osmotic pressure ($\pi$) can be calculated from the height difference ($h$) in the column: $\pi = \rho g h$ Given: $\rho$ (density of solution) = 1000 kg m$^{-3}$ $g$ = 10 m s$^{-2}$ $h$ = 80.0 mm = $80.0 \times 10^{-3}$ m $\pi = (1000 \text{ kg m}^{-3}) \times (10 \text{ m s}^{-2}) \times (80.0 \times 10^{-3} \text{ m})$ $\pi = 800 \text{ Pa}$ Now, use the van't Hoff equation for osmotic pr

Question

Q.73. 20 g hemoglobin in a 1 L aqueous solution (A) at 300 K is separated from pure water by semi permeable membrane. At equilibrium the height of solution in a tube dipped in a solution (A) is found to be 80.0 mm higher than the tube dipped in water. The molar mass of hemoglobin is ______ kg mol. (Nearest integer) (Given : g = 10 ms, R = 8.3 kPa dmKmol, density of solution = 1000 kg m)

FirstInTest · Accepted Answer

The osmotic pressure ($\pi$) can be calculated from the height difference ($h$) in the column:
$\pi = ho g h$
Given:
$ho$ (density of solution) = 1000 kg m$^{-3}$
$g$ = 10 m s$^{-2}$
$h$ = 80.0 mm = $80.0 	imes 10^{-3}$ m

$\pi = (1000 	ext{ kg m}^{-3}) 	imes (10 	ext{ m s}^{-2}) 	imes (80.0 	imes 10^{-3} 	ext{ m})$
$\pi = 800 	ext{ Pa}$

Now, use the van't Hoff equation for osmotic pressure:
$\pi = CRT$
Where:
$C$ = molar concentration (mol L$^{-1}$ or mol dm$^{-3}$)
$R$ = 8.3 kPa dm$^3$K$^{-1}$mol$^{-1}$ = $8.3 	imes 10^3$ Pa dm$^3$K$^{-1}$mol$^{-1}$ (since 1 kPa = 1000 Pa)
$T$ = 300 K

We need to find the molar mass (M) of hemoglobin. The concentration $C$ is given by:
$C = \frac{	ext{mass of solute}}{	ext{molar mass}} 	imes \frac{1}{	ext{volume of solution}}$
Mass of solute (hemoglobin) = 20 g
Volume of solution = 1 L = 1 dm$^3$

So, $C = \frac{20}{M} 	ext{ mol dm}^{-3}$

Substitute into the van't Hoff equation:
$800 	ext{ Pa} = \left(\frac{20}{M}ight) 	ext{ mol dm}^{-3} 	imes (8.3 	imes 10^3 	ext{ Pa dm}^3	ext{K}^{-1}	ext{mol}^{-1}) 	imes (300 	ext{ K})$

$800 = \frac{20 	imes 8.3 	imes 10^3 	imes 300}{M}$

$M = \frac{20 	imes 8.3 	imes 10^3 	imes 300}{800}$
$M = \frac{20 	imes 8.3 	imes 3000}{800}$
$M = \frac{20 	imes 8.3 	imes 30}{8}$
$M = \frac{5 	imes 8.3 	imes 30}{2}$
$M = 5 	imes 8.3 	imes 15$
$M = 41.5 	imes 15$
$M = 622.5 	ext{ g/mol}$

The question asks for the molar mass in kg mol$^{-1}$.
$M = 622.5 	ext{ g/mol} = 0.6225 	ext{ kg/mol}$

The nearest integer to 0.6225 is 1. However, the solution provided in the image calculates $M = 62.25 	ext{ kg mol}^{-1}$ and gives the answer as 62. Let's recheck the units and calculation in the solution.

Solution's calculation:
$\pi = 1000 	imes 10 	imes 80 	imes 10^{-3} = 800 	ext{ Pa}$
$C = \frac{20/M}{10^{-3} 	ext{ m}^3}$ (This implies M is in g/mol and volume is in $m^3$, so $1L = 10^{-3} m^3$)
$R = 8.3 	ext{ kPa dm}^3	ext{K}^{-1}	ext{mol}^{-1} = 8.3 	imes 10^3 	ext{ Pa m}^3	ext{K}^{-1}	ext{mol}^{-1}$ (since 1 dm$^3$ = $10^{-3}$ m$^3$, so $8.3 	ext{ kPa dm}^3 = 8.3 	imes 10^3 	ext{ Pa} 	imes 10^{-3} 	ext{ m}^3 = 8.3 	ext{ Pa m}^3$)
This R value seems incorrect for $Pa m^3$. Let's use $R = 8.314 	ext{ J mol}^{-1}	ext{K}^{-1} = 8.314 	ext{ Pa m}^3	ext{mol}^{-1}	ext{K}^{-1}$. The given R is $8.3 	ext{ kPa dm}^3	ext{K}^{-1}	ext{mol}^{-1}$.

Let's use the given R value directly with consistent units.
$R = 8.3 	ext{ kPa dm}^3	ext{K}^{-1}	ext{mol}^{-1} = 8.3 	ext{ kPa L K}^{-1}	ext{mol}^{-1}$
$\pi = 800 	ext{ Pa} = 0.8 	ext{ kPa}$
$C = \frac{20}{M} 	ext{ mol L}^{-1}$

$0.8 	ext{ kPa} = \left(\frac{20}{M}ight) 	ext{ mol L}^{-1} 	imes (8.3 	ext{ kPa L K}^{-1}	ext{mol}^{-1}) 	imes (300 	ext{ K})$

$0.8 = \frac{20 	imes 8.3 	imes 300}{M}$

$M = \frac{20 	imes 8.3 	imes 300}{0.8}$
$M = \frac{20 	imes 8.3 	imes 3000}{8}$
$M = 5 	imes 8.3 	imes 3000$
$M = 124500 	ext{ g/mol} = 124.5 	ext{ kg/mol}$

This is different from the solution's 62.25 kg/mol. Let's re-examine the solution's calculation:
$M = \frac{20 	imes 8.3 	imes 300}{0.8} = 62250 	ext{ g/mol} = 62.25 	ext{ kg/mol}$.

My calculation matches the solution's numerical result if I use $R = 8.3 	ext{ J mol}^{-1}	ext{K}^{-1}$ and $\pi$ in Pa, and volume in $m^3$. Let's try that.
$R = 8.3 	ext{ J mol}^{-1}	ext{K}^{-1} = 8.3 	ext{ Pa m}^3	ext{mol}^{-1}	ext{K}^{-1}$
Volume = 1 L = $10^{-3}$ m$^3$
$C = \frac{20/M}{10^{-3}} 	ext{ mol m}^{-3}$

$\pi = CRT$
$800 = \frac{20/M}{10^{-3}} 	imes 8.3 	imes 300$
$800 = \frac{20 	imes 8.3 	imes 300}{M 	imes 10^{-3}}$
$M = \frac{20 	imes 8.3 	imes 300}{800 	imes 10^{-3}}$
$M = \frac{20 	imes 8.3 	imes 300}{0.8}$
$M = 62250 	ext{ g/mol} = 62.25 	ext{ kg/mol}$.

This matches the solution's intermediate result. The final answer is 62 (nearest integer). The discrepancy was in my initial unit conversion for R. The given R is $8.3 	ext{ kPa dm}^3	ext{K}^{-1}	ext{mol}^{-1}$. If we use $dm^3$ as L, then $R = 8.3 	ext{ kPa L K}^{-1}	ext{mol}^{-1}$. $\pi = 0.8 	ext{ kPa}$.
$0.8 = \frac{20}{M} 	imes 8.3 	imes 300$
$M = \frac{20 	imes 8.3 	imes 300}{0.8} = 62250 	ext{ g/mol} = 62.25 	ext{ kg/mol}$.

The calculation is consistent with the solution. The nearest integer to 62.25 is 62.

Related Questionsin Chemistry

More from JEE Main

Browse Engineering Questions by Subject

Practice More JEE Main Questions