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Molarity
Appears in
Concepts tested here
- Concentration of ions
- Mass of concentrated acid
All Questions
2013 NEET 1 question
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How many grams of concentrated nitric acid solution should be used to prepare 250 mL of 2.0 M $HNO_3$? The concentrated acid is 70% $HNO_3$.$M = \frac{W\times1000}{M_w\times V_{solution}(mL)} \Rightarrow 2 = \frac{W\times1000}{63\times250}$
$W = 31.5$ g of $HNO_3$
70% $HNO_3$ means 70 g $HNO_3$ in 100 g solution.
31.5 g $HNO_3$ is present in $\frac{100}{70}\times31.5 = 45$ g of solution.
2010 AIPMT-PRE 1 question
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25.3 g of sodium carbonate, $Na_2CO_3$, is dissolved in enough water to make 250 mL of solution. If sodium carbonate dissociates completely, molar concentration of sodium ion, $Na^+$, and carbonate ions, $CO_3^{2-}$, are respectively (Molar mass of $Na_2CO_3$ = 106 g $mol^{-1}$)Moles of $Na_2CO_3$ = $\frac{25.3}{106} = 0.2387$ mol
Molarity of $Na_2CO_3$ = $\frac{0.2387}{250}\times1000 = 0.955$ M
Complete dissociation: $Na_2CO_3 \rightarrow 2Na^+ + CO_3^{2-}$
$[Na^+] = 2\times0.955 = 1.910$ M
$[CO_3^{2-}] = 0.955$ M