A solution of protein (extracted from crabs) was prepared by dissolving 0.75 g in 125 cm3 of an aqueous solution. At 4°C an osmotic pressure rise of 2.6 mm of the solution was observed. Then molecular weight of protein is (Assume density of solution is 1.00 g/cm3)

Question

A solution of protein (extracted from crabs) was prepared by dissolving 0.75 g in 125 cm3 of an aqueous solution. At 4°C an osmotic pressure rise of 2.6 mm of the solution was observed. Then molecular weight of protein is (Assume density of solution is 1.00 g/cm3) (A) 9.4 × 105 (B) 5.4 × 105 (C) 5.4 × 1010 (D) 9.4 × 1010

Tardigrade Admin · Accepted Answer

Given h = 2.6 mm ∴ π=hd g =(2.6/10)×1×980 dyne / cm2 Also, π=(w/Vm) RT ∴ (2.6×1×980/10)=(0.75×8.314×107×277/125× m) m=5.4×105

Q. A solution of protein (extracted from crabs) was prepared by dissolving $0.75 g$ in $125 c m^{3}$ of an aqueous solution. At $4^{\circ} C$ an osmotic pressure rise of $2.6 mm$ of the solution was observed. Then molecular weight of protein is (Assume density of solution is $1.00 g / c m^{3}$ )

$9.4 \times 1 0^{5}$

$5.4 \times 1 0^{5}$

$5.4 \times 1 0^{10}$

$9.4 \times 1 0^{10}$

Given $h = 2.6 mm$
$∴ π = h d g = \frac{2.6}{10} \times 1 \times 980$ dyne / $c m^{2}$
Also, $π = \frac{w}{Vm} RT$
$∴ \frac{2.6 \times 1 \times 980}{10} = \frac{0.75 \times 8.314 \times 1 0 ^{7} \times 277}{125 \times m}$
$m = 5.4 \times 1 0^{5}$

Q. A solution of protein (extracted from crabs) was prepared by dissolving 0.75g in 125cm3 of an aqueous solution. At 4∘C an osmotic pressure rise of 2.6mm of the solution was observed. Then molecular weight of protein is (Assume density of solution is 1.00g/cm3)

9.4×105

5.4×105

5.4×1010

9.4×1010