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JEE Mains · Chemistry · STD 11 - 1. Some basic concept of chemistry

The \(NaNO _{3}\) weighed out to make \(50\, mL\) of an aqueous solution containing \(70.0 \,mg \,Na ^{+}\) per \(mL\) is \(...... \,g\). (Rounded off to the nearest integer) [Given : Atomic weight in \(g\, mol ^{-1}- Na : 23\); \(N : 14 ; O : 16]\)

  1. A \(19\)
  2. B \(13\)
  3. C \(17\)
  4. D \(10\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(13\)

Step-by-step Solution

Detailed explanation

\(Na ^{+}\) present in \(50\, ml\) \(=\frac{70\, mg }{1\, ml } \times 50\, ml =3500 mg =3.5\, gm\) moles of \(Na ^{+}=\frac{3.5}{23}=\) moles of \(NaNO _{3}\) weight of \(NaNO _{3}=\frac{3.5}{23} \times 85=12.993\, gm\)
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