JEE MAIN - Chemistry (2021 - 16th March Morning Shift - No. 19)
A 6.50 molal solution of KOH (aq.) has a density of 1.89 g cm$$-$$3. The molarity of the solution is ____________ mol dm$$-$$3. (Round off to the Nearest Integer).
[Atomic masses : K : 39.0 u; O : 16.0 u; H : 1.0 u]
[Atomic masses : K : 39.0 u; O : 16.0 u; H : 1.0 u]
Answer
9
Explanation
$$m = {{1000 \times M} \over {1000 \times d - M \times {M_{solute}}}}$$
$$6.5 = {{1000 \times M} \over {1890 - M \times 56}}$$
$$ \Rightarrow $$ $$12285 - 364M = 1000M$$
$$ \Rightarrow $$ $$1364M = 12285$$
$$ \Rightarrow $$ $$M = 9$$
$$6.5 = {{1000 \times M} \over {1890 - M \times 56}}$$
$$ \Rightarrow $$ $$12285 - 364M = 1000M$$
$$ \Rightarrow $$ $$1364M = 12285$$
$$ \Rightarrow $$ $$M = 9$$
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