What weight (in gram) of CuSO4⋅ 5H2O must be taken to make 0.5 litre of 0.01 M Cu ion solution? Express your answer by rounding-off to the nearest integer value. (Cu = 63 . 5 , S = 32 , O = 16 , H = 1)

Question

What weight (in gram) of CuSO4⋅ 5H2O must be taken to make 0.5 litre of 0.01 M Cu ion solution? Express your answer by rounding-off to the nearest integer value. (Cu = 63 . 5 , S = 32 , O = 16 , H = 1) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

We know, The molecular weight of CuSO4.5H2O = 249.5 g/mol Weight of CuSO4.5H2O = 249.5 g/mol× 0.5L× 0.01mol/L=1.2475 g≈ 1 g

Q. What weight (in gram) of $C u S O_{4} \cdot 5 H_{2} O$ must be taken to make $0.5$ litre of $0.01 M C u$ ion solution?
Express your answer by rounding-off to the nearest integer value. $(C u = 63.5, S = 32, O = 16, H = 1)$

We know,
The molecular weight of $C u S O_{4} .5 H_{2} O$ = $249.5 g / m o l$
Weight of $C u S O_{4} .5 H_{2} O$ = $249.5 g / m o l \times 0.5 L \times 0.01 m o l / L = 1.2475 g \approx 1 g$

Q. What weight (in gram) of CuSO4​⋅5H2​O must be taken to make 0.5 litre of 0.01MCu ion solution? Express your answer by rounding-off to the nearest integer value. (Cu=63.5,S=32,O=16,H=1)

We know, The molecular weight of CuSO4​.5H2​O = 249.5g/mol Weight of CuSO4​.5H2​O = 249.5g/mol×0.5L×0.01mol/L=1.2475g≈1g

Q. What weight (in gram) of $C u S O_{4} \cdot 5 H_{2} O$ must be taken to make $0.5$ litre of $0.01 M C u$ ion solution?
Express your answer by rounding-off to the nearest integer value. $(C u = 63.5, S = 32, O = 16, H = 1)$

We know,
The molecular weight of $C u S O_{4} .5 H_{2} O$ = $249.5 g / m o l$
Weight of $C u S O_{4} .5 H_{2} O$ = $249.5 g / m o l \times 0.5 L \times 0.01 m o l / L = 1.2475 g \approx 1 g$