How many grams of solute are needed to prepare each of the following solutions?
A. 1.000 L of 0.180 M NaCl
B. 250.0 mL of 0.150 M CuSO4
C. 500.00 mL of 0.355 M CH3OH
mass= molarity*volume*molmassSolute
figure the mole mass of each, then do the math.
To calculate the grams of solute needed for each of the given solutions, we will use the formula:
grams of solute = (Molarity) x (volume in liters) x (molar mass of the solute)
First, let's find the molar mass of each solute:
A. NaCl: The molar mass of Na + Cl = 22.990 g/mol + 35.453 g/mol = 58.443 g/mol
B. CuSO4: The molar mass of Cu + (S x 4) + (O x 4) = 63.546 g/mol + (32.07 g/mol x 4) + (16.00 g/mol x 4) = 159.609 g/mol
C. CH3OH: The molar mass of C + (H x 4) + O + H = 12.011 g/mol + (1.008 g/mol x 4) + 16.00 g/mol + 1.008 g/mol = 32.042 g/mol
Now, we can calculate the grams of solute needed for each solution:
A. 1.000 L of 0.180 M NaCl:
grams of NaCl = (0.180 mol/L) x (1.000 L) x (58.443 g/mol) = 10.52 g
B. 250.0 mL of 0.150 M CuSO4:
Since the given volume is in milliliters, we need to convert it to liters:
Volume in liters = 250.0 mL / 1000 mL/L = 0.250 L
grams of CuSO4 = (0.150 mol/L) x (0.250 L) x (159.609 g/mol) = 5.97 g
C. 500.00 mL of 0.355 M CH3OH:
Again, let's convert the given volume to liters:
Volume in liters = 500.00 mL / 1000 mL/L = 0.500 L
grams of CH3OH = (0.355 mol/L) x (0.500 L) x (32.042 g/mol) = 5.70 g
Therefore, the grams of solute needed for each solution are:
A. 10.52 grams of NaCl
B. 5.97 grams of CuSO4
C. 5.70 grams of CH3OH
To calculate the number of grams of solute needed to prepare each solution, you need to use the equation:
moles of solute = concentration (in M) x volume (in L)
Then, you can use the equation:
grams of solute = moles of solute x molar mass of solute
where the molar mass is the mass of one mole of the solute.
Let's calculate the grams of solute for each case:
A. 1.000 L of 0.180 M NaCl:
First, calculate the moles of NaCl:
moles of NaCl = 0.180 M x 1.000 L = 0.180 moles
Next, calculate the molar mass of NaCl:
sodium (Na) has a molar mass of 22.99 g/mol
chlorine (Cl) has a molar mass of 35.45 g/mol
molar mass of NaCl = 22.99 g/mol + 35.45 g/mol = 58.44 g/mol
Finally, calculate the grams of NaCl:
grams of NaCl = 0.180 moles x 58.44 g/mol = 10.52 grams
Therefore, 10.52 grams of NaCl are needed to prepare 1.000 L of 0.180 M NaCl solution.
B. 250.0 mL of 0.150 M CuSO4:
First, convert mL to L:
Volume (in L) = 250.0 mL ÷ 1000 = 0.250 L
Next, calculate the moles of CuSO4:
moles of CuSO4 = 0.150 M x 0.250 L = 0.0375 moles
The molar mass of CuSO4 is:
copper (Cu) has a molar mass of 63.55 g/mol
sulfur (S) has a molar mass of 32.06 g/mol
oxygen (O) has a molar mass of 16.00 g/mol
molar mass of CuSO4 = (63.55 g/mol) + (32.06 g/mol) + (16.00 g/mol x 4) = 159.61 g/mol
Finally, calculate the grams of CuSO4:
grams of CuSO4 = 0.0375 moles x 159.61 g/mol = 5.99 grams
Therefore, 5.99 grams of CuSO4 are needed to prepare 250.0 mL of 0.150 M CuSO4 solution.
C. 500.00 mL of 0.355 M CH3OH:
First, convert mL to L:
Volume (in L) = 500.00 mL ÷ 1000 = 0.500 L
Next, calculate the moles of CH3OH:
moles of CH3OH = 0.355 M x 0.500 L = 0.1775 moles
The molar mass of CH3OH is:
carbon (C) has a molar mass of 12.01 g/mol
hydrogen (H) has a molar mass of 1.01 g/mol
oxygen (O) has a molar mass of 16.00 g/mol
molar mass of CH3OH = (12.01 g/mol) + (1.01 g/mol x 4) + 16.00 g/mol = 32.04 g/mol
Finally, calculate the grams of CH3OH:
grams of CH3OH = 0.1775 moles x 32.04 g/mol = 5.68 grams
Therefore, 5.68 grams of CH3OH are needed to prepare 500.00 mL of 0.355 M CH3OH solution.