The molar mass of "=Ba(OH)_2"
"Molarity=" "Moles\\over Liter"
"= 171.33g\/ mol"
We need to turn "35 ml" of "0.200moles Ba(OH)_2" into its equivalent mass
This is done by multiplying;
"0.200\\times 171.34\\over35ml" "= 0.979g BaOH"
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