You visited us 1 times! Enjoying our articles? Unlock Full Access!

Molality

QUESTION 2.26...

Question

If the density of some lake water is $1.25 g m L^{- 1}$ and contains 92 g of $N a^{+}$ ions per kg of water, calculate the molarity of $N a^{+}$ ions in the lake.

Open in App

Solution

Step 1:
Mass of sodium ions = 92 g
Molar mass of sodium ions = 23 g/mol
Number of moles of sodium ions = $\frac{92 g}{23 g {m o l}^{- 1}} = 4$
Mass of water = 1 kg
Volume = $\frac{m a s s}{d e n s i t y} = \frac{1000 g}{1.25 g {m L}^{- 1}} = 0.8 L$
Step 2:
As we know, molarity = $\frac{N u m b e r o f m o l e s}{v o l u m e o f s o l u t i o n}$
Molarity of $N a^{+}$ ions (m) $= \frac{N u m b e r o f m o l e s N a^{+} i o n s}{V o l u m e o f s o l u t i o n i n l i t r e s}$
$∴ M o l a r i t y = \frac{4 m o l}{0.8 L} = 5 m o l L^{- 1}$ or $5 M$

Suggest Corrections

196

Similar questions

If the density of some lake water is 1.25 g mL⁻¹ and contains 92 g of Na⁺ ions per kg of water, calculate the molality of Na⁺ ions in the lake.

N a^{+}

N a^{+}

1.25

m L^{- 1}

92

N a^{+}

N a^{+}

N a

= 23

m o l^{- 1}

92

N a^{+} i o n s

N a^{+}

k g^{- 1}

N a^{+}

N a^{+}

m o l k g^{- 1}

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program