Answer to Question #175700 in General Chemistry for Hong rong Zheng

Question #175700

What volume (in mL) of 1.2 M HNO3 is needed to contain 51.8 g HNO3?


1
Expert's answer
2021-03-26T05:45:08-0400

Molar Mass of HNO3 = 63.01g/mol

1.2M × 63.01g/mol = 75.612 g

51.8g/63.01g/mol = 0.8221 M

= (1.2× 75.612)/0.8221

= 110g


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