Van't Hoff factors of equimolal solutions of sodium chloride, barium chloride and glucose in water are

Question

Van't Hoff factors of equimolal solutions of sodium chloride, barium chloride and glucose in water are (A) 2, 3, 0 respectively (B) 2, 3, 6 respectively (C) 2, 3, 4 respectively (D) 2, 3, 1 respectively

Tardigrade Admin · Accepted Answer

NaCl ( aq ) longrightarrow Na ( aq )++ Cl ( aq )- van't Hoff factor, i=2 BaCl 2( aq ) longrightarrow Ba ( aq )2++2 Cl ( aq )-, i=3 Glucose does not undergo dissociation or association hence, i=1.

Q. Van't Hoff factors of equimolal solutions of sodium chloride, barium chloride and glucose in water are

$2, 3, 0$ respectively

$2, 3, 6$ respectively

$2, 3, 4$ respectively

$2, 3, 1$ respectively

$N a C l_{(a q)} ⟶ N a_{(a q)}^{+} + C l_{(a q)}^{-}$
van't Hoff factor, $i = 2$
$B a C l_{2 (a q)} ⟶ B a_{(a q)}^{2 +} + 2 C l_{(a q)}^{-}, i = 3$
Glucose does not undergo dissociation or association hence, $i = 1$ .

Q. Van't Hoff factors of equimolal solutions of sodium chloride, barium chloride and glucose in water are

2,3,0 respectively

2,3,6 respectively

2,3,4 respectively

2,3,1 respectively

NaCl(aq)​⟶Na(aq)+​+Cl(aq)−​ van't Hoff factor, i=2 BaCl2(aq)​⟶Ba(aq)2+​+2Cl(aq)−​,i=3 Glucose does not undergo dissociation or association hence, i=1.

$2, 3, 0$ respectively

$2, 3, 6$ respectively

$2, 3, 4$ respectively

$2, 3, 1$ respectively

$N a C l_{(a q)} ⟶ N a_{(a q)}^{+} + C l_{(a q)}^{-}$
van't Hoff factor, $i = 2$
$B a C l_{2 (a q)} ⟶ B a_{(a q)}^{2 +} + 2 C l_{(a q)}^{-}, i = 3$
Glucose does not undergo dissociation or association hence, $i = 1$ .