Question

A 36.8 g/L solution of K4 [Fe(CN)6g) (mol. wt. is
368) is isotonic with a 5% solution of a non-
volatile (non-electrolyte) solute. The molecular
weight of the solute will be
(1) 300
(2) 500
(3) 50
(4) 100​

Answer 1

The molecular weight of the solute will be 500

Explanation:-

Isotonic solutions are those solutions which have the same osmotic pressure. If osmotic pressures are equal at the same temperature, concentrations must also be equal.

$\pi =iCRT$

$\pi$ = osmotic pressure

i = vant hoff factor  ( no of ions produced on dissociation)

C= concentration

R= solution constant

T = temperature

Thus: $i\times C_{K_4[Fe(CN)_6]}=i\times C_{non electrolyte}$ where concentration is in molarity.

For :  ${K_4[Fe(CN)_6]}$ : 36.8 g is dissolved in 1 L of solution.

For non electrolyte: 5 g is dissolved in 100 ml of solution.

$Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution in ml}}$

$C_{K_4[Fe(CN)_6]}=5\times \frac{36.8}{368\times 1L}=0.1M$

$C_{non electrolyte}=1\times \frac{5g\times 1000}{M\times 100}$

$0.1=\frac{5\times 1000}{M\times 100}$

M=500g/mol

The molecular weight of the solute will be 500

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