question 1
The moles of C12H22O11 that are in 7.5 L of a 5.8 M C12H22O11 solution is 43.5 moles
<u><em>calculation</em></u>
<u><em> </em></u>moles = molarity x volume
volume = 7.5 L
molarity = 5.8 M or 5.8 mol/l
moles is therefore = 7.5 l x 5.8 mol/l = 43.5 moles
question 2
% percent =( mas of the solute/ mass of the solution) x 100
mass of solute= ?
mass of the solution = 375 g
% mass = 85 % = 85/100
let the mass of solute be represented by Y
therefore y/375 = 85/100
y=(375 x 85)/100 = 318.75 g
moles = mass/molar mass
= 318.75 g/ 42.4= 7.52 moles
Question 3
The grams of copper metal is 857.25 g
calculate the moles HCl
= molarity x volume = 4.50 L x 6 .00 mol/l= 27 moles
By use of mole ratio of Cu: HCl which is 1:2 the moles of Cu = 27/2 =13.5 moles
mass = moles x molar mass
= 13.5 moles x 63.5 g/mol = 857.25 g
question 4
Complete description of how to make 3.5 L of 2.0 M NaCl solution from a stock of 6.0 M NaCl is that measure 1.17 L of 6.0 M NaCl, Dissolve it to 3.5 L to make 2.0 M NaCl
<u><em>calculation</em></u>
M1V1 = M2V2
m1= 2.0 M
V1=3.5 L
M2= 6.0 M
V2=?
v2 = M1V1/M2
= (3.5 L x 2.0 M)/ 6.0 M= 1.17 L
explanation : measure 1.17 L of 6.0M NaCl and then dissolve it to 3.5 L to make 2.0 M NaCl
question 5
The number of grams of stock solution that is in 90.0 % is HCl by mass that would be needed to make 175 g is 77.78 grams
calculation
This is calculated using M1C1= M2C2 formula
M1 = ?g
C1 =90.0%
M1= 175 g
C1= 40.0 %
make M1 the subject of the formula
M1= M2C2/C1
M1 is therefore =( 175 g x 40) / 90 = 77.78 g
Question 6
The grams of silver metal is 127 g
% mass = mass of solute/ mass of solution
% mass = 80% = 80/100
mass of solute (AgNO3)= ?
mass of solution = 250 g
let the mass of solute be represented by Y
therefore Y/ 250 = 80/100
Y= (250 x 80) /100 =200 g of AgNO3
moles = mass/molar mass
moles of AgNO3 = 200 g/ 169.87 g/mol =1.178 moles
The mole ratio of AgNO3: Ag is 2:2=1:1 therefore the moles of Ag= 1.178 moles
mass= moles x molar mass
= 1.178 moles x 107.87 g/mol =127.07 g
