Question

A decrease of 1.2°C in freezing point took place on dissolving 7.5 gram of substance in 75 gram of water. find the molecular weight of substance. (molal depression constant for water k'f= 1.86°C).
​

Answer 1

Molecular weight of substance is 155 g/mol

Explanation:-

Depression in freezing point = $1.2^oC$

Mass of solute = 7.5 g

Mass of water = 75 g = 0.075 kg    (1 kg = 1000 g)

Formula used :

$\Delta T_f=K_f\times m\\\\\Delta T_f=i\times K_f\times\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of water in Kg}}$

where,

$\Delta T_f$ = change in freezing point = $1.2^0C$

$K_f$ = freezing point constant = $1.86^0C/kgmole$

m = molality

Now put all the given values in this formula, we get

$1.2=1.86\times \frac{7.5g}{\text{Molar mass of solute}\times 0.075kg}$

$\text{Molar mass of solute}=155g/mol$

The molecular weight of substance is 155 g/mol

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Answer 2

Answer:

Explanation:-

Depression in freezing point = 1.2^oC1.2

o

C

Mass of solute = 7.5 g

Mass of water = 75 g = 0.075 kg (1 kg = 1000 g)

Formula used :

\begin{gathered}\Delta T_f=K_f\times m\\\\\Delta T_f=i\times K_f\times\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of water in Kg}}\end{gathered}

ΔT

f

=K

f

×m

ΔT

f

=i×K

f

×

Molar mass of solute×Mass of water in Kg

Mass of solute

where,

\Delta T_fΔT

f

= change in freezing point = 1.2^0C1.2

0

C

K_fK

f

= freezing point constant = 1.86^0C/kgmole1.86

0

C/kgmole

m = molality

Now put all the given values in this formula, we get

1.2=1.86\times \frac{7.5g}{\text{Molar mass of solute}\times 0.075kg}1.2=1.86×

Molar mass of solute×0.075kg

7.5g

\text{Molar mass of solute}=155g/molMolar mass of solute=155g/mol

The molecular weight of substance is 155 g/mol