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Calculate the amount of solute dissolved in 612 grams of water at $30^{\circ} \mathrm{C}$ if molar mass of solute is $342 \mathrm{~g} \mathrm{~mol}^{-1}$ (Relative vapour pressure lowering is 0.025 and molar mass of water $18 \mathrm{~g} \mathrm{~mol}^{-1}$ )
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The correct answer is:
290.7 gram
$\frac{\Delta P}{P_A^{\circ}}=x_B \simeq \frac{n_B}{n_A}$
$\frac{W_B}{342} \times \frac{18}{612}=0.025$
$W_B=290.7$
$\frac{W_B}{342} \times \frac{18}{612}=0.025$
$W_B=290.7$
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