# The relative lowering of vapour pressure produced by dissolving 71.5 g of a substance in 1000 g of water is 0.00713. The molecular weight of the substance will be Option 1) $18.0$ Option 2) $342$ Option 3) $60$ Option 4) $180$

V Vakul

As we learned

Relative lowering of vapour pressure -

Relative lowering of vapour pressure on adding a non-volatile solute in a colligative properties.

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$\frac{P^{o}-P_{s}}{P^{0}}=\frac{\frac{w}{m}}{\frac{w}{m}+\frac{W}{M}}\: or\; 0.00713=\frac{71.5/m}{\frac{71.5}{m}+\frac{1000}{18}}$

$m=180$

Option 1)

$18.0$

Option 2)

$342$

Option 3)

$60$

Option 4)

$180$

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