# Question fb855

Mar 15, 2016

$0.20 M$

#### Explanation:

1. The equation is
$C a C {O}_{3} \to C a O + C {O}_{2}$
2. Find the formula masses of $C {O}_{2}$ and $C a C {O}_{3}$
$C a C {O}_{3}$=$100$ grams per mole
$C a = 1$x$40 = 40$
$C = 1$x$12 = 12$
$O = 3$x$16 = 48$
$C {O}_{2}$=$44$ grams per mole
$C = 1$x$12 = 12$
$O = 2$x$16 = 32$
3. Find the mass of pure marble;
$m a s s p u r e = \left(a c t u a l m a s s\right) \left(p u r i t y\right)$
mass pure=((10g)(0.50)
$m a s s p u r e = 5 g$
4. Convert mass $C a C {O}_{3}$ to mole $C a C {O}_{3}$
$5 \cancel{g C a C {O}_{3}}$x (1 mol CaCO_3)/(100 cancel(g CaCO_3)#
$0.05 m o l C a C {O}_{3}$
5. Find the mol $C {O}_{2}$; referring to the equation it's observed that both compounds have same number of mole; thus
$C {O}_{2} = 0.05 m o l$
6. Convert $250 m l$ to Li; $0.25 L i$
7. Therefore, to find the Molarity(M) use the formula:
$M = \frac{n . m o l}{L i s o l u t i o n}$
$M = \frac{0.05 m o l}{0.25 L i}$
8. $M = 0.20 \frac{m o l}{L i}$ or $0.20 M$