Chemistry question bank

Solving Chemistry question bank is an important habit which will help in clearing the particular concepts in this section(Chemistry question bank) i will provide you the amazing – amazing , mind booster Chemistry question bank. the question will be related all class including compitive exam too. and you can visit the here for concepts and for video lecture visit here.

Q) What is Chemical formula of sodium ascorbate.?

Chemical Formula of Sodium ascorbate = $C_6 H_7 Na O_6$ .

Molecular mass of sodium ascorbate = 198.106 g/mol

Sodium ascorbate is mineral salt of ascorbic acid(which is full of vitamins). and sodium ascorbate salt is also known as mineral ascorbate.

Q) A sample of hydrogen gas at a pressure of 2.00 atm and at temperature of 35 °C occupies a volume of 800 mL .if the gas is compressed at constant temperature until its pressure is 3.00 atm some of the gas sample will be …………… mL

Given data :-

initial pressure of hydrogen gas = 2.00 atm
Temperature = 35 °C
initial volume = 800 mL = 0.800 L
final pressure = 3.00 atm

Formula used :-

PV = nRT

where :-

P = pressure
V = Volume
n = number of moles
R = ideal gas constant
T = temperature

if Number of moles(n), R and temperature(T) is constant then

P₁V₁ = P₂V₂

where,

$P_1$ = initial pressure

$V_1$ = initial volume

$P_2$ = final pressure

$V_2$ = final volume

Calculation:-

$P_1 V_1 =P_2 V_2$

$V_2 = \frac{P_1V_1}{P_2}$ = $\frac{2.00 atm \times 0.800L}{3.00atm}$ = $\frac {1.6}{3.00} L$ = 0.5333 L

$V_2$ = 0.5333 L = 533.3 mL

Volume of sample of gas = 533.33 mL

Q) what amount of excess reagent remains when 5.0 g zinc reacts with 4.0 g phosphorus.

excess reagent = the chemical compound present in excess amount in the chemical reaction.

limiting reagent = the chemical compound which present in lesser amount in the chemical reaction.

data provided in the question is :-

amount of zinc = 5.0 g
amount of phosphorus = 4.0 g
molecular mass of Zinc(Zn) = 63.39 g/mol
molecular mass of phosphorus(P) = 30.97 g/mol
molecular mass of $Zn{_{3}}P{_{2}}$ = 258.1 g/mol

chemical reaction :-

$3Zn + 2P \rightarrow Zn{_{3}}P{_{2}}$

3×63.39 = 190.17 g of zinc reacts with 2×30.97 = 61.97 g of phosphorus then,

5.0 g of zinc will reacts with $\frac{61.94}{190.17}$ ×5.0g of phosphorus

= 1.628 g of phosphorus

5.0 g of zinc will reacts with 1.628 g of phosphorus.

excess reagent = phosphorus

limiting reagent = zinc

so the excess reagent remains in the reaction is = 4.0 g – 1.628 g = 2.372 g

excess reagent remains in the reaction is 2.372 g

Q) What volume of a 0.5 M potassium hydroxide solution is required to neutralize 20.0 mL of a 1.0 mL of hydrobromic acid solution. …………………. mL potassium hydroxide

sol.:-

potassium hydroxide :- KOH
hydrobromic acid :- HBr
1000 mL = 1 L
data provided in the question is :-

concentration of potassium hydroxide = 0.5 M = $M$
concentration of hydrobromic acid = 1.0 M = $M$
volume of hydrobromic acid = 20.0 mL = 0.020 L = $V{_{2}}$

formula used :-
$V1 = MV$
calculation :-
let the volume of potassium hydroxide is $V{_{1}}$
then,
from the formula used :-
$V1 = MV$
$0.5 M \times V_1= 1.0 M \times 0.020L$

$V_1 = \frac {1.0 M \times 0.020 L}{0.5M}$

$V_1$ = 0.04 L = 40 mL

so the volume of potassium hydroxide is 40 mL

Q) Calculate number of molecules contained in 15.0 g of silicon dioxide ?

Q) Calculate the mass in grams aluminum iodide( $AlI_3$ ) when 20.0 g of aluminum(Al) reacted completely with iodine( $I_2$ ) according to this reaction:- $2Al + 3 I_2 \rightarrow 2AlI_3$
a) 456.7 g
b) 318.8 g
c) 323.0 g
d) 167.7 g

data provided in the question is :-

mass of aluminum(Al) = 20.0 g

Reaction is :- $2Al + 3 I_2 \rightarrow 2AlI_3$

data require is :-

atomic mass of aluminum = 27 g/mol
atomic mass of iodine(I) = 127 g/mol
molecular mass of iodine ( $I_2$ ) = $2 \times 127 \frac{g}{mol}$
molecular mass of $AlI_3$ = 408 g/mol

then from the reaction :- 2 × 27 g = 54 g of Al reacts with iodine to form 2 × 408 g = 816 g of $AlI_3$
20.0 g Al will reacts with iodine and form

$\frac{816 g}{54 g} \times 20.0 g$ = 302.22 g
therefore 20.0 g of aluminum(Al) will completely reacts with iodine( $I_2$ ) to form 302.22 g of aluminum iodide( $AlI_3$ )