Neet Solved Paper 2013 Question 9
Question: How many grams of concentrated nitric acid solution should be used to prepare 250 mL of $ 2\text{.0M} $ $ HN{O_3} $ The concentrated acid is 70% $ HN{O_3} $ .
Options:
A) 45.0 g cone. $ HN{O_3} $
B) 90.0 g conc. $ HN{O_3} $
C) 70.0 g cone. $ HN{O_3} $
D) 54.0 g conc. $ HN{O_3} $
Show Answer
Answer:
Correct Answer: A
Solution:
Given, morality of solution = 2 Volume of solution $ =250,mL=\frac{250}{1000}=\frac{1}{4}L $
Molar mass of $ HNO _3=1+14+3\times 16=63\ g\ mo{l^{-1}} $ $ \because $
Molarity $ =\frac{weight,of,HNO _3}{mass,of,HNO _3\times volume,of,solution,(L)} $
$ \therefore $ Weight of $ HNO _3= $ morality $ \times $ mol.
Mass $ =2\times 63\times \frac{1}{4}g=31.5g $ It is the weight of $ 100%,HNO _3. $ But the given acid is $ 70%HNO _3. $
$ \therefore $ Its weight $ =31.5\times \frac{100}{70}g=45g $
BETA
