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Question

How many grams of the concentrated nitric acid solution is required to prepare $$250mL$$ of $$2.0M$$ nitric acid solution? (Concentration of $$H{NO}_{3}$$ is $$70\%$$)


A
45g
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B
90g
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C
70g
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D
54g
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Solution

The correct option is A $$45g$$
The required number of moles= $$=(\dfrac { 250 }{ 1000 } )×2=0.5$$moles
                               (no. of moles=molarity × volume)
so,required mass of $$HN{ O }_{ 3 }=0.5×63=31.5grams$$ 
           (given mass=no. of moles × molar mass)
Given,
70 grams of $$HN{ O }_{ 3 }$$ are present in 100 grams of the solution,
so,1 gram will be present in $$\dfrac { 100 }{ 70 } $$ grams of solution
hence 31.5 grams will be present in $$\dfrac { 100 }{ 70 } ×31.5 $$ grams of solution.
so the amount of concentrated nitric acid solution used is 45 grams

Chemistry

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