JEE MAIN - Chemistry (2021 - 27th July Morning Shift - No. 14)
The density of NaOH solution is 1.2 g cm$$-$$3. The molality of this solution is _____________ m. (Round off to the Nearest Integer)
[Use : Atomic mass : Na : 23.0 u, O : 16.0 u, H : 1.0 u, Density of H2O : 1.0 g cm$$-$$3]
[Use : Atomic mass : Na : 23.0 u, O : 16.0 u, H : 1.0 u, Density of H2O : 1.0 g cm$$-$$3]
Answer
5
Explanation
Consider 1 litre solution
mass of solution = (1.2 $$\times$$ 1000)g = 1200 gm
Neglecting volume of NaOH
Mass of water = 1000 gm
$$\Rightarrow$$ Mass of NaOH = (1200 $$-$$ 1000)gm = 200 gm
$$\Rightarrow$$ Moles of NaOH = $${{200g} \over {50g/mol}} = 5$$ mol
$$\Rightarrow$$ molality = $${{5mol} \over {1kg}} = 5$$ m
mass of solution = (1.2 $$\times$$ 1000)g = 1200 gm
Neglecting volume of NaOH
Mass of water = 1000 gm
$$\Rightarrow$$ Mass of NaOH = (1200 $$-$$ 1000)gm = 200 gm
$$\Rightarrow$$ Moles of NaOH = $${{200g} \over {50g/mol}} = 5$$ mol
$$\Rightarrow$$ molality = $${{5mol} \over {1kg}} = 5$$ m
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