JEE MAIN - Chemistry (2019 - 10th January Evening Slot - No. 16)
The amount of sugar (C12H22O11) required to prepare 2L of its 0.1 M aqueous solution is :
17.1 g
34.2 g
68.4 g
136.8 g
Explanation
Molarity = $${{{{(n)}_{solute}}} \over {{V_{solution}}(in\,\,lit)}}$$
0.1 = $${{wt./342} \over 2}$$
wt (C12H22O11) = 68.4 gram
0.1 = $${{wt./342} \over 2}$$
wt (C12H22O11) = 68.4 gram
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