The 25°C evaporation of a 100 cm\(^{3}\) solution of K\(_{2}\)CO\(_{3}\) to dryness gave 14g of the salt. What is the solubility of K\(_{2}\)CO\(_{3}\) at 25°C? [K\(_{2}\)CO\(_{3}\) = 138]
a
0.01 mol dm\(^{-3}\)
b
0.101 mol dm\(^{-3}\)
c
1.01 mol dm \(^{-3}\)
d
10.0 mol dm\(^{-3}\)
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Discussions (4)
TOMstar
4 years ago
Mass of k2CO3=14g
Vol. of solvent=100cm³
Solubility of K2CO3= x
No in moles=mass/Rmm=14/138
=0.101mole
100cm³ will dissolve 0.101mole of solute
1000cm³ will dissolve x
X= 0.101*1000/100
=1.01mol/dm³
Sel_ee
1 year ago
I'm pretty sure the formula is mass/molar mass ×1000/volume
that'll be 14/138 × 1000/100 and the answer will be 10.0
