To solve question 7, we can use the molarity equation:
M1V1 = M2V2
M1 is the molarity of the stock solution (2.0 mol/dm³)
V1 is the volume of stock solution needed
M2 is the desired molarity (0.15 mol/dm³)
V2 is the final volume of the diluted solution (500 cm³)
Plugging in the given values, we get:
2.0 mol/dm³ * V1 = 0.15 mol/dm³ * 500 cm³
Simplifying, we get:
V1 = (0.15 mol/dm³ * 500 cm³) / 2.0 mol/dm³
V1 = 7.5 cm³
Therefore, (B) 7.5 cm³ is the correct answer for question 7.
To solve question 8, we can use the same molarity equation:
M1V1 = M2V2
M1 is the molarity of the stock solution (unknown)
V1 is the volume of stock solution needed
M2 is the desired molarity (0.1 mol/dm³)
V2 is the final volume of the diluted solution (250 cm³)
Since the molarity and volume are the same for both NaOH and water (assuming we are diluting the stock solution with water), the ratio between moles and volume will be the same:
M1V1 = M2V2
M1 = (M2V2) / V1
M1 = (0.1 mol/dm³ * 250 cm³) / V1
To calculate the mass of NaOH required, we need to use the molar mass of NaOH, which is 22.99 g/mol + 16.00 g/mol + 1.008 g/mol = 39.998 g/mol.
The equation for calculating moles is:
moles = mass / molar mass
Rearranging the equation, we get:
mass = moles * molar mass
Using this equation, we can calculate the mass of NaOH required:
mass = (M2V2) / V1 * molar mass
mass = (0.1 mol/dm³ * 250 cm³) / V1 * 39.998 g/mol
We cannot solve this equation without knowing the value of V1, the volume of stock solution needed.
Therefore, we cannot determine the correct answer for question 8 without more information.
question 7 and 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) .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
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