0.471 m means 0.471 mols Co(NO3)2 in 1 kg solvent.
How many g Co(NO3)2 is 0.471 mols? That's g = mols x molar mass = ? = approx 86 grams Co(NO3)2
Total mass of the solution then is approx 86 + 1000 = approx 1086g total.
% = (g solute/total grams solution)*100 = ?
An aqueous solution of cobalt(II) nitrate has a concentration of 0.471 molal.
The percent by mass of cobalt(II) nitrate in the solution is %.
4 answers
0.471 m means 0.471 mols Co(NO3)2 in 1 kg solvent.
How many g Co(NO3)2 is 0.471 mols? That's g = mols x molar mass = ? = approx 86 grams Co(NO3)2
Total mass of the solution then is approx 86 + 1000 = approx 1086g total.
% = (g solute/total grams solution)*100 = ?
i don't get this what you mean????
How many g Co(NO3)2 is 0.471 mols? That's g = mols x molar mass = ? = approx 86 grams Co(NO3)2
Total mass of the solution then is approx 86 + 1000 = approx 1086g total.
% = (g solute/total grams solution)*100 = ?
i don't get this what you mean????
? means I assume you can do what comes before the ?
g = mols x molar mass.
You want to know grams Co(NO3)2. You know mols Co(NO3)2. You can look up the molar mass Co(NO3)2.
So to obtain grams (which you must do to solve the problem) you take mols (0.471) from above and multiply by molar mass (which I didn't bother to addup since you can do that). mols x molar mass = grams but I didn't multiply for you.
g = mols x molar mass.
You want to know grams Co(NO3)2. You know mols Co(NO3)2. You can look up the molar mass Co(NO3)2.
So to obtain grams (which you must do to solve the problem) you take mols (0.471) from above and multiply by molar mass (which I didn't bother to addup since you can do that). mols x molar mass = grams but I didn't multiply for you.
An aqueous solution of manganese(II) chloride has a concentration of 0.392 molal.
The percent by mass of manganese(II) chloride in the solution is
%.
The percent by mass of manganese(II) chloride in the solution is
%.