Question

Number of ion of Al+3
present in 250 ml of water of
a solution having concentration 3 molal
(1) 6.023 x
${10}^{23}$

(2) 4.5 x
${10}^{23}$

(3) 9.03 x
${10}^{23}$

(4) 12.046 x
${10}^{23}$
​

Answer 1

Answer:

4.5*10^23

Explanation:

molality = moles of solute / mass of solvent in kg

mass of solvent = 250g = 0.25kg

molality =3

               3 = x / 0.25

               x = 0.75

ions = 0.75 * 6.022*10^23

        = 4.5*10^23

Answer 2

Answer:

that's what i'm saying

accounting to Bernoulli's theorem

P1 + rho.g.h1 + 1/2mv1² = P2 +rho.g.h2 +1/2mv2²

p1 = p°

where P2 = (p° +rho.g.h)

V1 = 0

h2=0

so, P° + rho.g.h1 = P2 + 1/2mv2²

where i m doing wrong.... :(