Find the molality of the thiosulphate $(N a_{2} S_{2} O_{3})$ solution, if the density of a $3 M$ thiosulphate $(N a_{2} S_{2} O_{3})$ solution is $1.2 g/mL$ .

4.13 m

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2.57 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3.85 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5.46 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $4.13 m$
$3 M$ solution means 3 moles of sodium thiosulphate is present in $1 L$ of the solution.
Mass of ( $N a_{2} S_{2} O_{3}$ ) in solution = $3 \times m o l a r m a s s$
Molar mass of ( $N a_{2} S_{2} O_{3} = 158 g/mol$ )
Mass of solute = $3 \times 158 = 474 g$
Since, Density = $\frac{m a s s}{v o l u m e}$
Mass of solution $= 1.2 g/mL \times 1000 m L = 1200 g$
$Mass of solvent = mass of solution - mass of solute$
$= 1200 - 474 = 726 g$
$M o l a l i t y (m) = \frac{number of moles of solute}{mass of solvent in kg} \times 1000$
$M o l a l i t y = \frac{3}{726} \times 1000 = 4.13 m$

Suggest Corrections

Similar questions

(N a_{2} S_{2} O_{3})

3

(N a_{2} S_{2} O_{3})

1.58

N a^{+}

S_{2} O_{3}^{2 -}

3

(N a_{2} S_{2} O_{3})

1.58

N a_{2} S_{2} O_{3}

(N a_{2} S_{2} O_{3})

(N a_{2} S_{2} O_{3})

1.25 g / m L

N a^{+}

S_{2} O_{3}^{2 -}

Join BYJU'S Learning Program