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L M 21. What volume (ml) of a 5.45 M lead nitrate solution must be a to 20.7 ML M solution of lead nitrate? VPL NO 3 = ? a) 0
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Answer #1

M, - 5.us M V,=? U2 = 82007 M2 = 1041 M So here no. of moles of solute is constant in both case so MU, = M2 Uz 5.45 VI = 820.

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