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100 g of solution containing 50 g of NaOH.
If you think to a 2 M solution dissolve 80 g NaOH in 1 L demineralized water at 20 0C.
a solution of 0.1M NaOH has been prepared.How would you make 1L of the same solution
Given: 27 mL of NaOH, 0.45M; 20 mL HCI Need: M of HCI 27 ml NaOH*(1 L NaOH/1000mL NaOH)*(0.45M NaOH/1L NaOH)*(1mole HCI/1 mole NaOH)=0.012 0.012/0.02=0.607 M HCI (or rounded 0.61 M HCI)
We need to know the Molarity (or Molality or formality) of both the acid and the NaOH solution in order to answer this question.
0.01 molar
0.26
This solution contain 26,3 g NaOH.
The answer is 0,625 moles.
It is possible only if you evaporate the water.
The concentration of this solution (in NaOH) is 40 g/L.
standardization of NaoH