What is the molality of an aqueous solution, whose relative lowering of vapor pressure is $0.1$ ?

2.01 m

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2.75 m

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3.25 m

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Solution

The correct option is C $n o n e o f t h e a b o v e$
relative lowering of vapour pressure = $\frac{P^{0} - P}{P^{0}}; M o l a r m a s s o f H_{2} O = 18$
$\frac{P^{0} - P}{p^{0}} = m * \frac{M o l a r m a s s}{1000} => 0.1 = m * \frac{18}{1000}$
$m = \frac{0.1}{18} * 1000 = 5.55$

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What is the molality of an aqueous solution, whose relative lowering of vapor pressure is 0.1?

The correct option is C none of the aboverelative lowering of vapour pressure = P0−PP0;MolarmassofH2O=18P0−Pp0=m∗Molarmass1000=>0.1=m∗181000m=0.118∗1000=5.55

What is the molality of an aqueous solution, whose relative lowering of vapor pressure is $0.1$ ?

The correct option is C $n o n e o f t h e a b o v e$
relative lowering of vapour pressure = $\frac{P^{0} - P}{P^{0}}; M o l a r m a s s o f H_{2} O = 18$
$\frac{P^{0} - P}{p^{0}} = m * \frac{M o l a r m a s s}{1000} => 0.1 = m * \frac{18}{1000}$
$m = \frac{0.1}{18} * 1000 = 5.55$