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NEET 2010
Exam
NEET
Year
2010
Subject
Chemistry
Chapter
Solutions

NEET 2010 · Chemistry · Solutions

25.3 ~g of sodium carbonate, Na_2 CO_3 is dissolved in enough water...

25.3 g25.3 \mathrm{~g} of sodium carbonate, Na2CO3\mathrm{Na}_2 \mathrm{CO}_3 is dissolved in enough water to make 250 mL250 \mathrm{~mL} of solution. If sodium carbonate dissociates completely, molar concentration of sodium ion, Na+\mathrm{Na}^{+}and carbonate ion, CO32−\mathrm{CO}_3^{2-} are respectively (Molar mass of Na2CO3=106 g mol−1\mathrm{Na}_2 \mathrm{CO}_3=106 \mathrm{~g} \mathrm{~mol}^{-1} )25.3 ~g of sodium carbonate, Na_2 CO_3 is dissolved in enough water to make 250 ~mL of solution. If sodium carbonate25.3 ~g of sodium carbonate, Na_2 CO_3 is dissolved in enough water to make 250 ~mL of solution. If sodium carbonate

Options

  1. A0.955M0.955 \mathrm{M} and 1.910M1.910 \mathrm{M}
  2. B1.910M1.910 \mathrm{M} and 0.955M0.955 \mathrm{M}
  3. C1.90M1.90 \mathrm{M} and 1.910M1.910 \mathrm{M}
  4. D0.477M0.477 \mathrm{M} and 0.477M0.477 \mathrm{M}

Correct answer

B. 1.910M1.910 \mathrm{M} and 0.955M0.955 \mathrm{M}

Step-by-step solution

 Molarity = number of moles of solute  volume of solution (in mL) ×1000=25.3×1000106×250=0.9547≈0.955M\begin{aligned}\text{ Molarity }& =\frac{\text{ number of moles of solute }}{\text{ volume of solution (in mL) }} \times1000 \\ & =\frac{25.3 \times 1000}{106 \times 250}=0.9547\approx0.955 \mathrm{M} \end{aligned}\mathrm{Na}_2 \mathrm{CO}_3inaqueoussolutionremainsdissociatedasin aqueous solution remains dissociated as25.3 ~g of sodium carbonate, Na_2 CO_3 is dissolved in enough water to make 250 ~mL of solution. If sodium carbonateSince,themolarityof\\ Since, the molarity of\mathrm{Na}_2 \mathrm{CO}_3isis0.955 \mathrm{M},themolarityof, the molarity of\mathrm{CO}_3^{2-}isalsois also0.955 \mathrm{M}andthatofand that of\mathrm{Na}^{+}isis2 \times 0.955=1.910 \mathrm{M}$
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