Question #27284

How many grams of beryllium chloride are needed to make 125mL of a 0.050 M solution

Expert's answer

The equation needed to calculate theconcentration is

C(M) = n(mol) / V(L)

C - concentration (M)

n - amount of BeCl_{2} (mol)

V - volume (L)

n(mol) = C(M) x V(L) = 0.050 x 0.125 = 6.25 x 10^-3 mol

The mass of BeCl_{2} is

m(g) = n(mol) x M_{W}(g/mol)

M_{W}(BeCl_{2}) = 9.01 + 2 x 35.5 = 80.01 g/mol

m(BeCl_{2}) = 6.25 x 10^-3 x 80.01 = 0.500 g

Answer: m(BeCl_{2}) = 0.500g

C(M) = n(mol) / V(L)

C - concentration (M)

n - amount of BeCl

V - volume (L)

n(mol) = C(M) x V(L) = 0.050 x 0.125 = 6.25 x 10^-3 mol

The mass of BeCl

m(g) = n(mol) x M

M

m(BeCl

Answer: m(BeCl

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