Answer:
Weight of solution produced = 5135 kg
Amount of water removed = 4865 kg
Explanation:
For the balance of mass, the incoming mass of sugar must be equal to the outgoing mass. So, the incoming mass (mi) is 38% of 10000 kg
mi = 0.38x10000 = 3800 kg
The outgoing mass (mo) must be 3800 kg, and it is 74% of the total mass (mt)
mo = 0.74xmt
0.74xmt = 3800
mt = 3800/0.74
mt = 5135 kg
This is the mass of solution produced.
The amount of water removed (wr) is the amount of water incoming (wi) less the amount of water outgoing (wo). Both will be the total mass less the mass of sugar :
wi = 10000 - 3800 = 6200 kg
wo = 5135 - 3800 = 1335 kg
wr = wi - wo
wr = 6200 - 1335
wr = 4865 kg
Answer:
C. 500 cm' of 1.0 mol dmº magnesium sulphate solution.
Explanation:
Let us look at each of the solutions individually;
CaCl2 has three particles
K2SO4 has three particles
MgSO4 has two particles
C2H5OH has only one particle
The number of moles of moles in 250 cm of 2.0 mol dm-3 potassium chloride is 250/1000 * 2 = 0.5 moles having two particles
Also; number of moles in 500 cm' of 1.0 mol dm-3 magnesium sulphate solution= 500/1000 * 1 = 0.5 moles having two particles
Answer:
30 moles of ethanol are needed to prepare a 25 m solution, using 1200 g of water
Explanation:
This a short problem of solutions:
Our solute is ethanol: C₂H₅OH
Our solvent is water-
25 m, is our solution's concentration. m means molality (moles of solute contained in 1kg of solvent).
Our solvent mass is 1200g. We convert them to kg
1200 g . 1kg / 1000g = 1.2 kg
m = mol/kg → mol = m . kg
mol = 25 mol/kg . 1.2kg →30 moles
