【求助】请教下列文字如何翻译？

"add 3.0 mL of 0.24 N hydrochloric acid while swirling vigorously"

0.24 N hydrochloric acid 是24当量吧？不知道怎么理解。

在强烈涡旋的情况下加入3.0 mL 0.24 N的盐酸。

同意这个意见，不过对于盐酸来说，0.24当量与0.24mol/L是等同的，通常情况下，test翻译成“供试品”更专业一些。也就是供试品溶液的制备。

感谢qinyil207、ruojun和mhq111111的热心帮助，
想想的确昨天理解有问题，马上去试试看
如果test 翻译成供试品，那么与之对应的之前有一个叫做standard preparation 翻译成对照品溶液是否也可以？（这个standard preparation是进行系统适应性以及最后进行计算的）
原文见下：

Standard preparation— Transfer 5.0 mL of acetone to a 1000-mL volumetric flask, dilute with water to volume, and mix (Solution A). Transfer 5.0 mL of methanol to a 1000-mL volumetric flask, dilute with water to volume, and mix (Solution B). Transfer 50.0 mL of Solution A and 5.0 mL of Solution B to a 500-mL volumetric flask, dilute with water to volume, and mix to obtain a solution having concentrations of acetone and methanol of 0.050% and 0.005% (v/v), respectively.

译文：
溶液A：取5.0ml丙酮于1000ml容量瓶中，加水稀释至刻度，摇匀。
溶液B：精密量取5.0ml甲醇于1000ml容量瓶中，加水稀释至刻度，摇匀。
测试溶液：精密量取50.0ml溶液A和溶液5.0ml溶液B至500ml容量瓶，加水稀释至刻度，则该溶液含有0.050%丙酮和0.005%甲醇。

Procedure— [ NOTE— Use peak areas where peak responses are indicated. ] Separately inject equal volumes (about 2 µL) of the Standard preparation and the Test preparation into the chromatograph, record the chromatograms, and measure the acetone and methanol peak responses. Calculate the percentages of acetone and methanol in the sample taken by the same formula:
15.8P ( r U / r S),
in which P is the percentage (v/v) of acetone or methanol in the Standard preparation; and r U and r S are the acetone or methanol peak responses of the Test preparation and the Standard preparation, respectively: not more than 0.7% of acetone and 0.1% of methanol are found.

测定：分别进样相同体积的2ul的测试溶液和供试品溶液于色谱体系中，记录色谱图，测量丙酮和甲醇的峰面积，按下式计算样品含丙酮和甲醇的峰面积百分比：15.8P ( r U / r S) ，式中，P为测试溶液中丙酮或甲醇的百分数，rU和rS为试验溶液和测试溶液中的峰面积。标准规定：丙酮不得过0.7%，甲醇不得过0.1%。

请问以上是否还有什么翻译不当之处？
敬请指出，谢谢！

Standard preparation:标准溶液的制备（我的理解）
金山词霸的解释是：标准制剂

Standard preparation:标准溶液的制备
Test preparation:测试（待测）溶液的制备

谢谢茅茅，译为待测溶液更通俗专业。对应的就译为标准溶液吧