The student needs to dilute the stock to get the proper concentration. And during the dilution, the molar number of solute will not change. So the volume of stock needed is 2*0.1/1.75=0.114 L. Then dilute to 2.00 L.
FeO·TiO₂
w(Ti)=M(Ti)/M(FeO·TiO₂)
w(Ti)=47.88g/mol/151.73g/mol=0.3156 (31.56%)
Answer:
% sodium= 13.6 % sodium
% carbon= 35.5 % carbon
% hydrogen= 4.7% hydrogen
% nitrogen = 8.3% nitrogen
% oxygen = 37.8 % oxygen
Explanation:
To find its percent composition means that we are to find to find the percentage of each of the constituents of the compound present.
The molar mass of monosodium glutamate is 169.11 gmol-1
Hence;
Percent of sodium= 23 gmol-1/169.11 gmol-1 × 100 = 13.6 % sodium
Percent of Carbon= 60 gmol-1/169.11 gmol-1 ×100 = 35.5 % carbon
Percent of hydrogen= 8/169.11 gmol-1 ×100 = 4.7% hydrogen
Percent nitrogen = 14/169.11 gmol-1 × 100 = 8.3% nitrogen
Percent oxygen = 64/169.11 gmol-1 ×100 = 37.8 % oxygen
C because we need materials in the lab