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The vapour pressure of a solution of 5g of non electrolyte in 100g of water at a particular temperature is 2 2985 ${Nm}^{- 2}$ . The vapour pressure of pure water at that temperature is 3000 ${Nm}^{- 2}$ .The molecular weight of the solute is

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Correct option is A

Vapour pressure of solution=P=2985 ${Nm}^{- 2}$

Vapour pressure of water $= P^{0} = 300 {Nm}^{- 2}$

Wt. of non electrolyte solute $= W_{1} = 5 g$

Wt. of water $= W_{2} = 100 g$

GMW of water $= M_{2} = 18 g$

GMW of solute $= M_{1} = ?$

According to Raoult’s law

$\frac{P^{0} - P}{P^{0}} = \frac{\frac{W_{1}}{M_{1}}}{\frac{W_{1}}{M_{1}} + \frac{W_{2}}{M_{2}}}$

$\frac{3000 - 2985}{3000} = \frac{\frac{5}{M_{1}}}{\frac{5}{M_{1}} + \frac{100}{18}}$

$\frac{15}{3000} = \frac{\frac{5}{M_{1}}}{\frac{90 + 100 M_{1}}{18 M_{1}}}$

$\frac{15}{3000} = \frac{5}{M_{1}} \times \frac{18 M_{1}}{(90 + 100 M_{1)}}$

$90 + 100 M_{1} = \frac{5 \times 18 \times 3000}{15}$

$100 M_{1} = \frac{15000 \times 18}{15} - 90$

$100 M_{1} = 18000 - 90$

$M_{1} = \frac{17910}{100}$

$M_{1} = 179.1 (Near By)$

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The vapour pressure of a solution of 5g of non electrolyte in 100g of water at a particular temperature is 2 2985 Nm-2 . The vapour pressure of pure water at that temperature is 3000 Nm-2.The molecular weight of the solute is