Question

how to prepare 0.1m caco3 in 1000 ml of distilled water if atomic mass of ca 40U

Answer 1

Molarity of CaCO3 = 0.1

Molarity= No. of moles/Volume of solution in water

No. of moles= given weight/molecular mass

Let given weight be x gm

Molecular Weight = 40+12+(16×3)

                              = 100 gm/mole

Put value in above formula

$0.1=\frac{x/100}{1}$

x=0.1×100

x=10 gm

Answer 2

By dissolving 10 g of $CaCO_3$ in 1000 ml of distilled water.

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n\times 1000}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in ml

moles of $CaCO_3=\frac{\text {given mass}}{\text {Molar Mass}}=\frac{xg}{100g/mol}$

Now put all the given values in the formula of molarity, we get

$0.1M=\frac{x\times 1000}{100\times 1000}$

$x=10g$

Therefore  0.1M $CaCO_3$ is prepatred by dissolving 10 g of $CaCO_3$ in 1000 ml of distilled water

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