When 1.04 g of BaCl2 is present in 105 g of solution the concentration of solution is

Question

When 1.04 g of BaCl2 is present in 105 g of solution the concentration of solution is (A) 0.104 ppm (B) 10.4 ppm (C) 0.0104 ppm (D) 104 ppm

Tardigrade Admin · Accepted Answer

Concentration in ppm =( text mass of solute / text mass of solution ) × 106 =( text mass of BaCl 2/ text mass of solution ) × 106 beginarrayl =(1.04 g /105 g ) × 106 =10.4 ppm endarray

Q. When $1.04 g$ of $B a C l_{2}$ is present in $1 0^{5} g$ of solution the concentration of solution is

$0.104 pp m$

$10.4 pp m$

$0.0104 pp m$

$104 pp m$

Concentration in ppm $= \frac{mass of solute}{mass of solution} \times 1 0^{6}$
$< b r / >= \frac{mass of B a C l _{2}}{mass of solution} \times 1 0^{6} < b r / >$
$< b r / > < b r / >= \frac{1.04 g}{1 0 ^{5} g} \times 1 0^{6} < b r / >= 10.4 pp m < b r / > < b r / >$

Q. When 1.04g of BaCl2​ is present in 105g of solution the concentration of solution is

0.104ppm

10.4ppm

0.0104ppm

104ppm