The molality of the solution is obtained as 0.63 m.
<h3>What is the freezing point?</h3>
The freezing point is the temperature at which the liquid is converted into solid.
We know that;
ΔT = 3.5° C
K = 1.86° C/m
i = 3
m = ?
Thus;
ΔT = K m i
m = ΔT/K i
m = 3.5° C/ 1.86° C/m * 3
m = 0.63 m
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The grams of glucose are needed to prepare 400g of a 2.00%(m/m) glucose solution g is calculated as below
=% m/m =mass of the solute/mass of the solution x100
let mass of solute be represented by y
mass of solution = 400 g
% (m/m) = 2% = 2/100
grams of glucose is therefore =2/100 = y/400
by cross multiplication
100y = 800
divide both side by 100
y= 8.0 grams
Answer:
Both of the studies said that the mass of the atom is centered in the nucleus, which is positive, and there are electrons (negative particles) orbiting it. So, Rutheford and Nagaoka discovered that the atom can be divisible and it has an empty space.
But, in the model of Nagaoka, the nucleus was huge, and for Rutherford, the nucleus was really small, and the mass was concentrated. By his experiment with the gold sheets, the theory was appropriated. That's why Rutherford is credited with the discovery of the nucleus. Nagaoka was incorrect in his suppositions.
Evaporation of the solution
