What volume of aqueous solution of NaOH that is `80%` by mass NaOH, contanis 0.4 mol of NaOH. Density of solution is `0.8 gm//ml`.
A. `8 ml`
B. 25 ml
C. 16 ml
D. 250 ml
Correct Answer - B
`0.5 mol HNO_(3)-=0.5 molH^(+)`
`H^(+)+OH^(-)toH_(2)O`, `DeltaH=-57.1 kJ`
Here `OH^(-)=0.3 mol`
Thus, `OH^(-)` is the limiting reactant
When `1` mol of `OH^(-)` ions are neutralised heat released` =...
Correct Answer - A
Assume 1 L of each solution
`[H_(3)O^(+)]` in solution of `pH =6` is `10^(-6)M`
`[H_(3)O^(+)]` in solution of `pH=3` is `10^(-3) M`
Total `[H_(3)O^(+)]=((10^(-3)+10^(-6))/(2))`
`=5.005xx10^(-4)`
`:. pH...
Calculation of molarity of the solution
Let V mL of each sample of `H_(2)SO_(4)` bemixed for 30% `H_(2)SO_(4)`
Mass of `H_(2)SO_(4)`=30g ,Mass of solution=100 g
`" Volume of solution"=("Mass of soulution")/("Density")=...
Correct Answer - a
Since `H_(3)PO_(3)` is dibasic, 0.1 M solution =- 0.2 N solution.
`underset(("Acid"))(N_(1)V_(1))=underset(("Base"))(N_(2)V_(2))`
`(0.2 N)xx20 mL)=(0.1N)xxV_(2)`
`V_(2)=((0.2N)xx(20mL))/((0.1N))=40 mL.`
Correct Answer - d
By definition, `3.6 M (3xx98=352.8 g) of H_(2)SO_(4)` are present in 1000 mL of solution.
29.0 g of acid are present in solution=100 g
325.8 of acid...
Correct Answer - D
Let `M_(0)` = mass of body in vacuum.
Apparent weight of the body in air = Apparent weight of standard weights in air implies Actual weight -...