Question 7 to 8
7. To prepare 500~cm^{3} of 0.15 mol/dm³ sulphuric acid from a stock solution of 2.0~mol/dm^{3} sulphuric acid. The volume of stock solution required is (A)37~cm^{3}(B) (B)37.5~cm^{3}(C)27.8~cm^{3}(D) 25.4~cm^{3}
8. The mass of NaOH required to prepare 250~cm^{3} of 0.1M NaOH is (A) 10g (B) 4g (C) 40g (D) 1g
7. To prepare 500 cm^3 of 0.15 mol/dm³ sulphuric acid:
We can use the formula:
C1V1 = C2V2
C1 = concentration of stock solution = 2.0 mol/dm^3
V1 = volume of stock solution required
C2 = desired concentration = 0.15 mol/dm^3
V2 = desired volume = 500 cm^3
Plugging in the values:
2.0V1 = 0.15 * 500
V1 = (0.15 * 500) / 2.0
V1 = 7.5 cm^3
So, the volume of stock solution required is approximately 7.5 cm^3.
Therefore, the correct option is (B) 7.5 cm^3.
8. To calculate the mass of NaOH required to prepare 250 cm^3 of 0.1M NaOH:
We can use the formula:
Molarity (M) = moles/volume (dm^3)
Moles = Molarity * Volume (dm^3)
Molarity = 0.1M
Volume = 250 cm^3
Converting cm^3 to dm^3:
Volume = 250/1000 = 0.25 dm^3
Moles = 0.1 * 0.25 = 0.025 moles
To calculate the mass, we need to know the molar mass of NaOH:
Na (sodium) = 22.99 g/mol
O (oxygen) = 16.00 g/mol
H (hydrogen) = 1.01 g/mol
Molar mass of NaOH = 22.99 + 16.00 + 1.01 = 40.00 g/mol
Mass = Moles * Molar mass
Mass = 0.025 * 40.00 = 1.00 g
Therefore, the correct option is (D) 1g.