A is 0.0950moldm^3 HCl B is a solution containing 13.50gdm^-3 of X2CO3.10H2O. Take the volume...

A is 0.0950moldm^3 HCl B is a solution containing 13.50gdm^-3 of X2CO3.10H2O. Take the volume of A to be 24.60cm^3 and volume of B to be 25.00cm^3, calculate
(i) concentration of B in moldm^-3
(ii)molar mass of 10H2O in gmol^-1
(iii)percentage by mass of X in X2CO3 [H=1, C=12, O=16]
the equation for the reaction is
2HCl(aq)+X2CO3.10H2O------->
2XC(aq)+11H2O+CO2(aq)?

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Answers (3)

Jonathan
1 week ago
(i)CaVa/CbVb= na/nb
0.0950*24.60/25*Cb=2/1
Cb= 0.0950*24.60/25*2
Cb= 0.0467moldm^-3
(ii) molar conc
= mass conc/molar mass
molar mass(MM)=
mass conc/ molar conc
=13.50/0.0467
MM = 2896gmol^-1
(iii) X2CO3.10H2O= 289
2x +(1*12)+(3*16)+10(18)= 289
2x +12+48+180=289
2x+240= 289
2x= 49
x= 24.5
MM of X2CO3=(24.5*2)+12+(16*3)
=109
(2x/X2CO3)* 100%
(2*24.5/109)*100%
=44.9%
=45%(approximately)
victory
1 year ago
(i)CaVa/CbVb= na/nb
0.0950*24.60/25*Cb=2/1
Cb= 0.0950*24.60/25*2
Cb= 0.0467moldm^-3
(ii) molar conc
= mass conc/molar mass
molar mass(MM)=
mass conc/ molar conc
=13.50/0.0467
MM = 2896gmol^-1
(iii) X2CO3.10H2O= 289
2x +(1*12)+(3*16)+10(18)= 289
2x +12+48+180=289
2x+240= 289
2x= 49
x= 24.5
MM of X2CO3=(24.5*2)+12+(16*3)
=109
(2x/X2CO3)* 100%
(2*24.5/109)*100%
=44.9%
=45%(approximately)
Gaby
10 months ago
(i)CaVa/CbVb= na/nb
0.0950*24.60/25*Cb=2/1
Cb= 0.0950*24.60/25*2
Cb= 0.0467moldm^-3
(ii) molar conc
= mass conc/molar mass
molar mass(MM)=
mass conc/ molar conc
=13.50/0.0467
MM = 2896gmol^-1
(iii) X2CO3.10H2O= 289
2x +(1*12)+(3*16)+10(18)= 289
2x +12+48+180=289
2x+240= 289
2x= 49
x= 24.5
MM of X2CO3=(24.5*2)+12+(16*3)
=109
(2x/X2CO3)* 100%
(2*24.5/109)*100%
=44.9%
=45%(approximately)
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