JEE MAIN - Chemistry (2021 - 31st August Morning Shift - No. 15)
The molarity of the solution prepared by dissolving 6.3 g of oxalic acid (H2C2O4.2H2O) in 250 mL of water in mol L$$-$$1 is x $$\times$$ 10$$-$$2. The value of x is _____________. (Nearest integer) [Atomic mass : H : 1.0, C : 12.0, O : 16.0]
Answer
20
Explanation
$$[{H_2}{C_2}{O_4}.2{H_2}O] = {{weight/{M_w}} \over {V(L)}}$$
$$ \Rightarrow x \times {10^{ - 2}} = {{6.3/126} \over {250/1000}}$$
$$x = 20$$
$$ \Rightarrow x \times {10^{ - 2}} = {{6.3/126} \over {250/1000}}$$
$$x = 20$$
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