Answer to Question #112835 in Chemistry for Joy

Question

1. calculate the number of moles of sodium carbonate present in 100cm³ of 2M solution
2.calculate the mass of sulphuric (iv)acid in 250cm3 of solution whose concentration is 0.25mole dm³
3. Calculate the number of moles present in 25cm³ of a 0.2M sodium hydroxide solution
Mass in grams present in 25cm³ of a 0.2M sodium hyroxide solution
4.calculate the number of moles of sodium chloride in 500cm³ of 3M sodium chloride solution
5.calculate the mass of calcium chloride required to make one litre of 0.2M calcium chloride
6.when 34.8g of potassium sulphate were dissolved in 500cm³ distilled water.calc a)the concentration of potassium sulphate b) molarity of the solution

Answer 1

1. The number of moles equals:

n = M × V,

where n - number of moles, M - molarity, V - volume.

n(Na₂CO₃) = 2 M × 100 cm³ = 2 mol/L × 0.1 L = 0.2 mol

2. The mass equals:

m = n × Mr = M × V × Mr

From here:

m(H₂SO₄) = 0.25 mol/L × 0.25 L × 98.1 g/mol = 6.13 g

3. n(NaOH) = M × V = 0.2 M × 0.025 L = 0.005 mol

m(NaOH) = n × Mr = 0.005 mol × 40 g/mol = 0.2 g

4. n(NaCl) = M × V = 3 M × 0.5 L = 1.5 mol

5. M = n / V = m / (Mr × V)

where M - molarity, n - number of moles, V - volume, Mr - molecular weight.

From here:

m(CaCl₂) = M × Mr × V = 0.2 mol/L × 111 g/mol × 1 L = 22.2 g

6. a) c = m / V

where c - concentration, m - mass, V - volume

From here:

c(K₂SO₄) = 34.8 g / 0.5 L = 69.6 g/L

b) M = n / V = m / (Mr × V)

where M - molarity, n - number of moles, V - volume, Mr - molecular weight.

From here:

M(K₂SO₄) = 34.8 g / (174 g/mol × 0.5 L) = 0.4 mol/L